Problem: Simplify the following expression and state the condition under which the simplification is valid. $k = \dfrac{-2p^3 + 6p^2 + 8p}{5p^2 + 20p - 160}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ k = \dfrac {-2p(p^2 - 3p - 4)} {5(p^2 + 4p - 32)} $ $ k = -\dfrac{2p}{5} \cdot \dfrac{p^2 - 3p - 4}{p^2 + 4p - 32} $ Next factor the numerator and denominator. $ k = - \dfrac{2p}{5} \cdot \dfrac{(p - 4)(p + 1)}{(p - 4)(p + 8)}$ Assuming $p \neq 4$ , we can cancel the $p - 4$ $ k = - \dfrac{2p}{5} \cdot \dfrac{p + 1}{p + 8}$ Therefore: $ k = \dfrac{ -2p(p + 1)}{ 5(p + 8)}$, $p \neq 4$